Gujarat BoardĀ GSEB Textbook Solutions Class 7 Maths Chapter 2 Fractions and Decimals InText Questions and Answers.
Gujarat Board Textbook Solutions Class 7 Maths Chapter 2 Fractions and Decimals InText Questions
Try These (Page 34)
Question 1.
Find:
(a) \(\frac { 2 }{ 7 }\) x 3
(b) \(\frac { 9 }{ 7 }\) x 6
(c) 3 x \(\frac { 1 }{ 8 }\)
(d) \(\frac { 13 }{ 11 }\) x 6
If the product is an improper fraction express it as a mixed fraction.
Solution:
(a) \(\frac { 2 }{ 7 }\) x 3
= \(\frac { 2Ć3 }{ 7 }\)
= \(\frac { 6 }{ 7 }\)
(b) \(\frac { 9 }{ 7 }\) x 6
= \(\frac { 9Ć6 }{ 7 }\)
= \(\frac { 54 }{ 7 }\)
= 7\(\frac { 5 }{ 7 }\)
(c) 3 x \(\frac { 1 }{ 8 }\)
= \(\frac { 3Ć1 }{ 8 }\)
= \(\frac { 3 }{ 8 }\)
(d) \(\frac { 13 }{ 11 }\) x 6
= \(\frac { 13Ć6 }{ 11 }\)
= \(\frac { 78 }{ 11 }\)
= 7\(\frac { 1 }{ 11 }\)
Question 2.
Represent pictorially: 2 x \(\frac { 2 }{ 5 }\) = \(\frac { 4 }{ 5 }\)
Solution:
or 2 x \(\frac { 2 }{ 5 }\) = \(\frac { 4 }{ 5 }\)
Try These (Page 34)
Question 1.
Find: (i) 5 x 2\(\frac { 3 }{ 7 }\)
(ii) 1\(\frac { 4 }{ 9 }\) x 6
Solution:
(i) 5 x 2\(\frac { 3 }{ 7 }\)
= 5 x \(\frac { 17 }{ 7 }\)
= \(\frac { 85 }{ 7 }\)
= 12\(\frac { 1 }{ 7 }\)
(ii) 1\(\frac { 4 }{ 9 }\) x 6
= \(\frac { 13 }{ 9 }\) x 6
= 
= \(\frac { 13Ć2 }{ 3 }\)
= \(\frac { 26 }{ 3 }\)
= 8\(\frac { 2 }{ 3 }\)
Try These (Page 35)
Question 1.
Can you tell, what is
(i) \(\frac { 1 }{ 2 }\) of 10?
(ii) \(\frac { 1 }{ 4 }\) of 16?
(iii) \(\frac { 2 }{ 5 }\) of 25?
Solution:
(i) \(\frac { 1 }{ 2 }\) of 10
= \(\frac { 1 }{ 2 }\) x 10
= \(\frac { 1Ć10 }{ 2 }\)
= 5
(ii) \(\frac { 1 }{ 4 }\) of 16
= \(\frac { 1 }{ 4 }\) x 16
= \(\frac { 1Ć16 }{ 4 }\)
= 4
(iii) \(\frac { 2 }{ 5 }\) of 25
= \(\frac { 2 }{ 5 }\) x 25
= \(\frac { 2Ć25 }{ 5 }\)
= 10
Try These (Page 39)
Question 1.
Fill in these boxes:
(i) \(\frac { 1 }{ 2 }\) x \(\frac { 1 }{ 7 }\) = \(\frac { 1Ć1 }{ 2Ć7 }\) = 
(ii) \(\frac { 1 }{ 5 }\) x \(\frac { 1 }{ 7 }\) = =
(iii) \(\frac { 1 }{ 7 }\) x \(\frac { 1 }{ 2 }\) = =
(iv) \(\frac { 1 }{ 7 }\) x \(\frac { 1 }{ 5 }\) = =
Solution:
Try These (Page 40)
Question 1.
Find: \(\frac { 1 }{ 3 }\) x \(\frac { 4 }{ 5 }\); \(\frac { 2 }{ 3 }\) x \(\frac { 1 }{ 5 }\)
Solution:
(i) \(\frac { 1 }{ 3 }\) x \(\frac { 4 }{ 5 }\) = \(\frac { 1Ć4 }{ 3Ć5 }\) = \(\frac { 4 }{ 15 }\)
(ii) \(\frac { 2 }{ 3 }\) x \(\frac { 1 }{ 5 }\) = \(\frac { 2Ć1 }{ 3Ć5 }\) = \(\frac { 2 }{ 15 }\)
Try These (Page 40)
Question 1.
Find: \(\frac { 8 }{ 3 }\) x \(\frac { 4 }{ 7 }\); \(\frac { 3 }{ 4 }\) x \(\frac { 2 }{ 3 }\)
Solution:
(i) \(\frac { 8 }{ 3 }\) x \(\frac { 4 }{ 7 }\) = \(\frac { 8Ć4 }{ 3Ć7 }\) = \(\frac { 32 }{ 21 }\)
(ii) \(\frac { 3 }{ 4 }\) x \(\frac { 2 }{ 3 }\) = \(\frac { 3Ć2 }{ 4Ć3 }\) = \(\frac { 1Ć1 }{ 2Ć1 }\) = \(\frac { 1 }{ 2 }\)
Think, Discuss and Write (Page 44)
Question 1.
(i) Will the reciprocal of a proper fraction be again a proper fraction?
(ii) Will the reciprocal of an improper fraction be again an improper fraction!
Solution:
(i) No, the reciprocal of a proper fraction is an improper fraction.
(ii) No, the reciprocal of an improper fraction is a proper fraction.
Now, we can say that
(a) 1 Ć· \(\frac { 1 }{ 2 }\)
= 1 x \(\frac { 2 }{ 1 }\)
= 1 x reciprocal of \(\frac { 1 }{ 2 }\)
(b) 3 Ć· \(\frac { 1 }{ 4 }\)
= 3 x \(\frac { 4 }{ 1 }\)
= 3 x reciprocal of \(\frac { 1 }{ 4 }\)
(c) 3 Ć· \(\frac { 1 }{ 2 }\) = _____ = _____
3 x \(\frac { 1 }{ 2 }\) = 3 x \(\frac { 2 }{ 1 }\) reciprocal of \(\frac { 1 }{ 2 }\)
And, 2 Ć· \(\frac { 3 }{ 4 }\) = 2 x reciprocal of \(\frac { 3 }{ 4 }\)
= 2 x \(\frac { 4 }{ 3 }\)
(d) 5 Ć· \(\frac { 2 }{ 9 }\) = 5 x _____ = _____
ā“ 5 Ć· \(\frac { 2 }{ 9 }\) = 5 x \(\frac { 9 }{ 2 }\) = 5 x reciprocal of \(\frac { 2 }{ 9 }\)
Remember:
I. When the product of two fractions is unity, then each is called the āreciprocal of the otherā.
II. When unity is divided by a fraction, then the quotient is the āreciprocalā of that fraction.
Try These (Page 45)
Question 1.
Find:
(i) 7 Ć· \(\frac { 2 }{ 5 }\)
(ii) 6 Ć· \(\frac { 4 }{ 7 }\)
(iii) 2 Ć· \(\frac { 8 }{ 9 }\)
Solution:
Try These (Page 45)
Question 1.
Find:
(i) 6 Ć· 5\(\frac { 1 }{ 3 }\)
(ii) 7 Ć· 2\(\frac { 4 }{ 7 }\)
Solution:
Try These (Page 45)
Question 1.
Find:
(i) \(\frac { 3 }{ 5 }\) Ć· \(\frac { 1 }{ 2 }\)
(ii) \(\frac { 1 }{ 2 }\) Ć· \(\frac { 3 }{ 5 }\)
(iii) 2\(\frac { 1 }{ 2 }\) Ć· \(\frac { 3 }{ 5 }\)
(iv) 5\(\frac { 1 }{ 6 }\) Ć· \(\frac { 9 }{ 2 }\)
Solution:
Try These (Page 50)
Question 1.
Find:
(i) 2.7 x 4
(ii) 1.8 x 1.2
(iii) 2.3 x 4.35
Solution:
(i) 2.7 x 4
āµ 27 x 4 = 108 and there is one digit to the right of the decimal point in 27.
ā“ 2.7 x 4 = 10.8
(ii) 1.8 x 1.2
āµ 18 x 12 = 216 and number of digits to the right of decimal point is (1 + 1), i.e. 2.
ā“ 1.8 x 1.2 = 2.16
(iii) 2.3 x 4.35
āµ 23 x 435 = 10005 and there are 1+2, i.e.
ā“ digits to the right of decimal point.
ā“ 2.3 x 4.35 = 10.005
Question 2.
Arrange the products obtained in Question 1 in descending order.
Solution:
The products are: 10.8, 2.16, 10.005.
Comparing 10.8 and 10.005, we have:
10 = 10, 8 > 0, i.e. 10.005 < 10.8
Here, the smallest number = 2.16
and, the largest number = 10.8
Thus, the required descending order is: 10.8, 10.005, 2.16.
Try These (Page 51)
Question 1.
Find:
(i) 0.3 x 10
(ii) 1.2 x 100
(iii) 56.3 x 1000
Solution:
(i) 0.3 x 10
āµ There is 1 zero in 10.
ā“ The decimal point is shifted to the right by 1 place.
Thus, 0.3 x 10 = 3
(ii) 1.2 x 100
āµ There are 2 zeros in 100.
ā“ The decimal point is shifted to the right by 2 places.
Thus, 1.2 x 100 = 120
(iii) 56.3 x 1000
āµ There are three zeros in 1000.
ā“ The decimal point is shifted to the right by 3 places.
Thus, 56.3 x 1000 = 56300
Try These (Page 53)
Question 1.
Find:
(i) 235.4 Ć· 10
(ii) 235.4 + 100
(iii) 235.4 Ć· 1000
Solution:
(i) 235.4 Ć· 10
Since, there is one zero in 10.
ā“The decimal point in the quotient is shifted to the left by one place.
ā“ 235.4 Ć· 10 = 23.54
(ii) 235.4 Ć· 100
Since, there are two zeros in 100
ā“ The decimal point in the quotient is shifted to the left by two places.
ā“ 235.4 Ć· 100 = 2.354
(iii) 235.4 Ć· 1000
Since, there are three zeros in 1000.
ā“ The decimal point in the quotient is shifted to the left by three places.
ā“ 235.4 Ć· 1000 = 0.2354
Try These (Page 53)
Question 1.
Find:
(i) 35.7 Ć· 3
(ii) 25.5 Ć· 3
Solution:
(i) 35.7 Ć· 3
Since, \(\frac { 357 }{ 3 }\) =119 and there is one digit in the decimal part of the given decimal number.
ā“ The decimal point is placed in the quotient after one digit from the right most digit.
ā“ 35.7 Ć· 3 = 11.9
(ii) 25.5 Ć· 3
Since, 255 Ć· 3 = 85 and there is one digit in the decimal part of the given decimal number.
ā“ The decimal is placed in the quotient after one digit from the right most digit.
ā“ 25.5 Ć· 3 = 8.5
Try These (Page 53)
Question 1.
Find:
(i) 43.15 Ć· 5
(ii) 82.44 Ć· 6
Solution:
(i) 43.15 Ć· 5
Since 4315 Ć· 5 = 863 and there are two digits in the decimal part of the given decimal number.
ā“ Place the decimal point in 863 such that there are two digits to its right.
ā“ 43.15 Ć· 5 = 8.63
(ii) 82.44 + 6
Since 8244 Ć· 6 = 1374 and there are two digits in the decimal part of the given decimal number.
ā“ Place the decimal point in 1374 such that there are two digits to its right.
ā“ 82.44 Ć· 6 = 13.74
Try These (Page 53)
Question 1.
Find:
(i) 15.5 Ć· 5
(ii) 126.35 Ć· 7
Solution:
(i) 15.5 Ć· 5
Since 155 Ć· 5 = 31 and there is one digit in the decimal part of the given decimal number.
ā“ Place the decimal point in 31 such that there is one digit to its right.
ā“ 15.5 Ć· 5 = 3.1
(ii) 126.35 Ć· 7
Since 12635 Ć· 7 = 1805 and there are two digits in the decimal part of the given decimal number.
ā“ Place the decimal point in 1805 such that there are two digits to its right.
ā“ 126.35 Ć· 7 = 18.05
Try These (Page 54)
Question 1.
Find:
(i) \(\frac { 7.75 }{ 0.25 }\)
(ii) \(\frac { 42.8 }{ 0.02 }\)
(iii) \(\frac { 5.6 }{ 1.4 }\)
Solution:
(i) \(\frac { 7.75 }{ 0.25 }\)
since, 7.75 = \(\frac { 775 }{ 100 }\) and 0.25 = \(\frac { 25 }{ 100 }\)
ā“ 7.75 Ć· 0.25 = \(\frac { 775 }{ 100 }\) Ć· \(\frac { 25 }{ 100 }\)
= \(\frac { 775 }{ 100 }\) x \(\frac { 100 }{ 25 }\)
= \(\frac { 775 }{ 25 }\)
= 31
ā“ \(\frac { 7.75 }{ 0.25 }\) = 31
(ii) \(\frac { 42.8 }{ 0.02 }\)
since, 42.8 = \(\frac { 428 }{ 10 }\) and 0.25 = \(\frac { 2 }{ 100 }\)
ā“ 42.8 Ć· 0.25 = \(\frac { 428 }{ 100 }\) Ć· \(\frac { 2 }{ 100 }\)
= \(\frac { 428 }{ 10 }\) x \(\frac { 100 }{ 2 }\)
= \(\frac { 4280 }{ 25 }\)
= 2140
ā“ \(\frac { 42.8 }{ 0.02 }\) = 2140
(iii) \(\frac { 5.6 }{ 1.4 }\)
since, 5.6 = \(\frac { 56 }{ 10 }\) and 1.4 = \(\frac { 14 }{ 100 }\)s
ā“ 5.6 Ć· 1.4 = \(\frac { 56 }{ 10 }\) Ć· \(\frac { 14 }{ 10 }\)
= \(\frac { 56 }{ 10 }\) x \(\frac { 10 }{ 14 }\)
= \(\frac { 56 }{ 14 }\)
= 4
ā“ \(\frac { 5.6 }{ 1.4 }\) = 4